Answer
$dv_{}=0.0508\text{ m}^{3}\text{/kg}$
$\Delta v=0.0508\text{ m}^3\text{/kg}$
Work Step by Step
( $a$ ) The changes in $T$ and $P$ can be expressed as $$
\begin{aligned}
& d T \cong \Delta T=(305-300) \mathrm{K}=5 \mathrm{~K} \\
& d P \cong \Delta P=(96-100) \mathrm{kPa}=-4 \mathrm{kPa} \end{aligned}
$$ The ideal gas relation $P v=R T$ can be expressed as $v=R T / P$. Note that $R$ is a constant and $v=v(T, P)$. Applying the total differential relation and using average values for $T$ and $P$, $$
\begin{aligned}
d v & =\left(\frac{\partial v}{\partial T}\right)_P d T+\left(\frac{\partial v}{\partial P}\right)_T d P-\frac{R d T}{P}-\frac{R T d P}{P^2} \\
& =\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)\left(\frac{5 \mathrm{~K}}{98 \mathrm{kPa}}-\frac{(302.5 \mathrm{~K})(-4 \mathrm{kPa})}{(98 \mathrm{kPa})^2}\right) \\
& =\left(0.01464 \mathrm{~m}^3 / \mathrm{kg}\right)+\left(0.03616 \mathrm{~m}^3 / \mathrm{kg}\right) \\
& =0.0508 \mathrm{~m}^3 / \mathrm{kg}
\end{aligned}
$$ (b) Using the ideal gas relation at each state, $$
\begin{aligned}
& v_1=\frac{R T_1}{P_1}=\frac{\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)(300 \mathrm{~K})}{100 \mathrm{kPa}}=0.8610 \mathrm{~m}^3 / \mathrm{kg} \\
& v_2=\frac{R T_2}{P_2}=\frac{\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)(305 \mathrm{~K})}{96 \mathrm{kPa}}=0.9118 \mathrm{~m}^3 / \mathrm{kg}
\end{aligned}
$$ Thus, $$
\Delta v=v_2-v_1=0.9118-0.8610=0.0508 \mathrm{~m}^3 / \mathbf{k g}
$$
